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Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1
Question
Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree. 4tan^2 θ+tanθ=−4tanθ−1
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Mathematics
3 years
2021-08-28T06:28:01+00:00
2021-08-28T06:28:01+00:00 1 Answers
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Answer:
135 and 315degrees
Step-by-step explanation:
Given the expression
4tan²θ+tanθ=−4tanθ−1
Equate to zero
4tan²θ+tanθ+4tanθ+1 = 0
4tan²θ+5tanθ+1 = 0
Let x = tanθ
4x²+5x+1 = 0
Factorize;
4x²+4x+x+1 = 0
4x(x+1)+1(x+1) = 0
4x+1 = 0 and x+1 = 0
4x = -1 and x = -1
x = -1/4 and -1
Since x = tanθ
-1 =tanθ
θ = arctan(-1)
θ = -45degrees
Since tan is negative in the 2nd and fourth quadrant;
In the second quadrant
θ = 180 – 45 = 135degrees
In the fourth quadrant;
θ = 360 – 45 = 315degrees
Hence the required angles are 135 and 315degrees